Binomial Expansion : Lesson

From Pascal’s Triangle, the coefficients for the expansion of \((x + a)^5 \) are:
\(^5C_0,\ ^5C_1,\ ^5C_2,\ ^5C_3,\ ^5C_4,\ ^5C_5 = 1,\ 5,\ 10,\ 10,\ 5,\ 1 \)

Using these coefficients:

\((x + 3)^5 = 1 \cdot x^5 + 5 \cdot x^4 \cdot 3 + 10 \cdot x^3 \cdot 3^2 + 10 \cdot x^2 \cdot 3^3 + 5 \cdot x \cdot 3^4 + 1 \cdot 3^5 \)

\(\quad = x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243 \)

\(\quad = x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243 \)

From Pascal’s Triangle, the coefficients for the expansion of \((a + b)^4 \) are:
\(^4C_0,\ ^4C_1,\ ^4C_2,\ ^4C_3,\ ^4C_4 = 1,\ 4,\ 6,\ 4,\ 1\)

Using these coefficients:

\((2x – 3)^4 = ^4C_0(2x)^4 + ^4C_1(2x)^3(-3) + ^4C_2(2x)^2(-3)^2 + ^4C_3(2x)(-3)^3 + ^4C_4(-3)^4\)

\(\quad = 1 \cdot 16x^4 + 4 \cdot 8x^3 \cdot (-3) + 6 \cdot 4x^2 \cdot 9 + 4 \cdot 2x \cdot (-27) + 1 \cdot 81\)

\(\quad = 16x^4 – 96x^3 + 216x^2 – 216x + 81\)

[a] Number of Terms

For the expansion of \((a + b)^n\) , the number of terms is $n+1$

So the expansion of \((2x + 3y)^5\)  has \(\quad 5 + 1 = 6\) terms.

[b] Full Expansion Using Pascal’s Triangle

Here, \(a = 2x\), \(b = 3y\), and \(n = 5\)

The number of terms in the expansion is: \(n + 1 = 6\)

Using Pascal’s Triangle, the coefficients are:
\(1,\ 5,\ 10,\ 10,\ 5,\ 1\)

Now apply the binomial expansion:

Now apply the binomial expansion:

\((2x + 3y)^5 = 1 \cdot (2x)^5 + 5 \cdot (2x)^4(3y) + 10 \cdot (2x)^3(3y)^2\)

\(\quad\quad\quad\quad +\ 10 \cdot (2x)^2(3y)^3 + 5 \cdot (2x)(3y)^4 + 1 \cdot (3y)^5\)

\((2x + 3y)^5 = 32x^5 + 240x^4y + 720x^3y^2 + 1080x^2y^3 + 810xy^4 + 243y^5\)

We apply the binomial expansion formula:

\((a + b)^n = \sum_{r=0}^{n} \ ^nC_r \cdot a^{n-r} \cdot b^r\)

Here,

\(a = 2x\), \(b = -\frac{3}{x}\), and \(n = 4\)

Using the binomial expansion:

\((2x – \frac{3}{x})^4 = \ ^4C_0 (2x)^4(-\frac{3}{x})^0 + \ ^4C_1 (2x)^3(-\frac{3}{x})^1 + \ ^4C_2 (2x)^2(-\frac{3}{x})^2 \)

\(\quad\quad\quad\quad\quad\quad\quad\quad +\ ^4C_3 (2x)^1(-\frac{3}{x})^3 + \ ^4C_4 (2x)^0(-\frac{3}{x})^4\)

Now simplify each term:

\(\ ^4C_0 (2x)^4(-\frac{3}{x})^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4\)

\(\ ^4C_1 (2x)^3(-\frac{3}{x})^1 = 4 \cdot 8x^3 \cdot (-\frac{3}{x}) = -96x^2\)

\(\ ^4C_2 (2x)^2(-\frac{3}{x})^2 = 6 \cdot 4x^2 \cdot \frac{9}{x^2} = 216\)

\(\ ^4C_3 (2x)^1(-\frac{3}{x})^3 = 4 \cdot 2x \cdot (-\frac{27}{x^3}) = -216/x^2\)

\(\ ^4C_4 (2x)^0(-\frac{3}{x})^4 = 1 \cdot 1 \cdot \frac{81}{x^4} = 81/x^4\)

Final Answer :  \((2x – \frac{3}{x})^4 = 16x^4 – 96x^2 + 216 – \frac{216}{x^2} + \frac{81}{x^4}\)

We are expanding: \((2x + 1)^7\)

The general term in a binomial expansion is: \(T_{r+1} =\ ^nC_r \cdot a^{n – r} \cdot b^r\)

Here, \(a = 2x\),  \(b = 1\),  \(n = 7\)

To find the 4th term, we use \(r = 3\) (since \(T_{r+1} = T_4\))

Now substitute into the formula:

\(T_4 =\ ^7C_3 \cdot (2x)^{7 – 3} \cdot (1)^3\)

Simplify each part:

\(\ ^7C_3 = 35\)

\((2x)^4 = 16x^4\)

\((1)^3 = 1\)

Now calculate:

\(T_4 = 35 \cdot 16x^4 \cdot 1 = 560x^4\)

Final Answer:   \({560x^4}\)

(a) In any binomial expansion of the form \((a + b)^n\), the number of terms is: \(n + 1\)

Here, \(n = 11\). \(\Rightarrow\) Number of terms = \(11 + 1 = {12}\)

(b) We are asked to find the term containing \(x^3\).

We are given the expansion:  \((x + 2)^{11}\)

The general term in the binomial expansion is given by:

\(T_{r+1} =\ ^{11}C_r \cdot x^{11 – r} \cdot 2^r\)

We want:

\(x^{11 – r} = x^3 \Rightarrow 11 – r = 3 \Rightarrow r = 8\)

Now substitute $r = 8$ into the general term:

\(T_9 =\ ^{11}C_8 \cdot x^{3} \cdot 2^8\)

\(= 165 \cdot x^3 \cdot 256 = \boxed{42240x^3}\)

Final Answer:

\(\boxed{42240x^3}\) is the term containing $x^3$ in the expansion of $(x + 2)^{11}$.

Solution

[a] Number of terms:

The number of terms in the expansion of \((a + b)^n\) is \(n + 1\).

So, for \(n = 5\), the number of terms is:  \({6\text{ terms}}\)

[b] Find the term containing \(x^4\):

We use the general term formula:

\(T_{r+1} =\ ^5C_r \cdot (2x^2)^{5 – r} \cdot (-\frac{3}{x})^r\)

Now simplify the powers of $x$:

\(x\text{-power} = 2(5 – r) – r = 10 – 3r\)

Set exponent equal to 4:
\(10 – 3r = 4 \Rightarrow r = 2\)

Now substitute \(r = 2\) into the formula:

\(T_3 =\ ^5C_2 \cdot (2x^2)^3 \cdot \left(-\frac{3}{x}\right)^2\)

\(= 10 \cdot 8x^6 \cdot 9x^{-2} = 10 \cdot 72x^4 = {720x^4}\)

Final Answer:

\({720x^4}\) is the term containing $x^4$ in the expansion of \((2x^2 – \frac{3}{x})^5\).

[a] Expand \((x + 2)^4 \)

We apply the binomial expansion:

\((x + 2)^4 = ^4C_0 x^4 + ^4C_1 x^3 \cdot 2 + ^4C_2 x^2 \cdot 2^2 + ^4C_3 x \cdot 2^3 + ^4C_4 \cdot 2^4\)

Now calculate each term:

\(= 1 \cdot x^4 + 4 \cdot x^3 \cdot 2 + 6 \cdot x^2 \cdot 4 + 4 \cdot x \cdot 8 + 1 \cdot 16\)

\(= x^4 + 8x^3 + 24x^2 + 32x + 16\)

So, \((x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16\)

[b] Expand \((x – 3)(x + 2)^4\)

From part (a), we already know:

\((x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16\)

Now multiply by \((x – 3)\):

\((x – 3)(x^4 + 8x^3 + 24x^2 + 32x + 16)\)

Use distributive property:

\(= x(x^4 + 8x^3 + 24x^2 + 32x + 16) – 3(x^4 + 8x^3 + 24x^2 + 32x + 16)\)

First distribute \(x\):

\(= x^5 + 8x^4 + 24x^3 + 32x^2 + 16x\)

Then distribute \(-3\):

\(= -3x^4 -24x^3 -72x^2 -96x -48\)

Now combine like terms:

\(= x^5 + (8x^4 – 3x^4) + (24x^3 – 24x^3) + (32x^2 – 72x^2) + (16x – 96x) – 48\)

\(= x^5 + 5x^4 – 40x^2 – 80x – 48\)

So, the first few terms in descending powers: \(x^5 + 5x^4 – 40x^2\)

[a] Number of terms

The number of terms in any binomial expansion of the form \((a + b)^n\) is \(n + 1\).

So for \((x^2 – 3x)^7\), the number of terms is: \(7 + 1 = 8\)

[b] Coefficient of \(x^{11}\)

Let us consider the general term in the expansion of \((x^2 – 3x)^7\):

\(T_{r+1} = ^7C_r (x^2)^{7 – r}(-3x)^r\)

Simplify each component:

\((x^2)^{7 – r} = x^{2(7 – r)} = x^{14 – 2r}\)

\((-3x)^r = (-3)^r x^r\)

So the term becomes:

\(T_{r+1} = ^7C_r (-3)^r x^{14 – 2r + r} = ^7C_r (-3)^r x^{14 – r}\)

To find the coefficient of \(x^{11}\):

We set the power of \(x\) equal to 11:

\(14 – r = 11 \quad \Rightarrow \quad r = 3\)

Substitute \(r = 3\) into the coefficient:

\(\text{Coefficient} = ^7C_3 \cdot (-3)^3 = 35 \cdot (-27) = -945\)

[c] Show that there is no constant term

From the general term:

\(T_{r+1} = ^7C_r (-3)^r x^{14 – r}\)

A constant term would occur when the power of \(x\) is zero:

\(14 – r = 0 \quad \Rightarrow \quad r = 14\)

But in this expansion, \(r\) can take values from 0 to 7 only.

Since \(r = 14\) is not within this range, there is no constant term in the expansion.

[a] Expand \((e + \frac{1}{e})^5\)

Using the binomial expansion:

\((e + 1/e)^5 = \ ^5C_0 \cdot e^5 + \ ^5C_1 \cdot e^3 + \ ^5C_2 \cdot e + \ ^5C_3 \cdot 1/e + \ ^5C_4 \cdot 1/e^3 + \ ^5C_5 \cdot 1/e^5\)

\(= e^5 + 5e^3 + 10e + \frac{10}{e} + \frac{5}{e^3} + \frac{1}{e^5}\)

[b] Expand \((e – \frac{1}{e})^5\)

Using alternating signs in binomial expansion:

\((e – 1/e)^5 = \ ^5C_0 \cdot e^5 – \ ^5C_1 \cdot e^3 + \ ^5C_2 \cdot e – \ ^5C_3 \cdot 1/e + \ ^5C_4 \cdot 1/e^3 – \ ^5C_5 \cdot 1/e^5\)

\(= e^5 – 5e^3 + 10e – \frac{10}{e} + \frac{5}{e^3} – \frac{1}{e^5}\)

[c] Using results from (a) and (b)

(i) \((e + \frac{1}{e})^5 + (e – \frac{1}{e})^5\)

\(= (e^5 + 5e^3 + 10e + \frac{10}{e} + \frac{5}{e^3} + \frac{1}{e^5}) + (e^5 – 5e^3 + 10e – \frac{10}{e} + \frac{5}{e^3} – \frac{1}{e^5})\)

\(= 2e^5 + 20e + \frac{10}{e^3}\)

(ii) \((e + \frac{1}{e})^5 – (e – \frac{1}{e})^5\)

\(= (e^5 + 5e^3 + 10e + \frac{10}{e} + \frac{5}{e^3} + \frac{1}{e^5}) – (e^5 – 5e^3 + 10e – \frac{10}{e} + \frac{5}{e^3} – \frac{1}{e^5})\)

\(= 10e^3 + \frac{20}{e} + \frac{2}{e^5}\)